國中數學,圓,切線,三角形,面積

2010-06-15 9:14 am

回答 (3)

2010-06-15 10:29 am
✔ 最佳答案

圖片參考:http://imgcld.yimg.com/8/n/AD04686329/o/151006150056813871800250.jpg

(Note: F為BE直線與圓的交點,請自行補上EF線段)
1. ∆PAB:∆PAD:∆PAM=PA:PD:PM
2. PA+PD=2PE (E為圓心O至AD之垂足)
3. (重點)PBEC四點共圓
因∆PBC之外接圓即OP之直徑之圓
又∠OEP=90度, so, E為∆PBC之外接圓上
故∠CPB=∠CBF
即(CF弧+FD弧-AC弧)/2=CF弧/2 or AC弧=DF弧
4. PA*PD=PB^2 (切割線性質)
5. 1/∆PAB+ 1/∆PAD= 2/∆PAM
<=> 1/PA+ 1/PD= 2/PM
<=> 2PE/PB^2= 2/PM
<=> PB/PM = PE/PM
<=> ∆PBM~∆PEB
<=> ∠PBM=∠PEB
<=> ∠PBA+∠ABM=∠EDB+∠EBD
<=> ∠ABM=∠EBD
<=> AC弧=DF弧
(Note:第5步驟倒敘即得證)

2010-06-15 03:05:30 補充:
5. 1/∆PAB+ 1/∆PAD= 2/∆PAM
<=> 1/PA+ 1/PD= 2/PM
<=> 2PE/PB^2= 2/PM
<=> PB/PM = PE/PB
<=> ∆PBM~∆PEB
<=> ∠PBM=∠PEB
<=> ∠PBA+∠ABM=∠EDB+∠EBD
<=> ∠ABM=∠EBD
<=> AC弧=DF弧

2010-06-15 23:44:19 補充:
To:老王
請問:圓對稱的調和性質,是什麼幾何定理?
我可是證明得很辛苦,感覺繞得很遠!

2010-06-16 18:58:58 補充:
謝謝老王!
2010-06-16 12:44 am
BC是P對圓O的極線
所以P、A、M、D是調和點列
1/PA+1/PD=2/PM
結論得證

2010-06-16 17:48:31 補充:
TO:天助
那三句話,都要證很久
有興趣參考一下
http://tw.myblog.yahoo.com/oldblack-wang/article?mid=1714
http://tw.myblog.yahoo.com/oldblack-wang/article?mid=1722
2010-06-15 10:29 pm
3.∠CPD=∠CBF


收錄日期: 2021-04-21 22:18:42
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